Draw an arc of the circle with center RRR and radius Q⁢R¯normal-¯QR\overline{QR} to find the pointSSS where it intersects the arc from the previous step such that S≠QSQS\neq Q.

....PPPQQQRRRSSS

7.

Draw the square P⁢Q⁢R⁢SPQRSPQRS.

....PPPQQQRRRSSS

This construction is justified because P⁢S=P⁢Q=Q⁢R=Q⁢SPSPQQRQSPS=PQ=QR=QS, yielding that P⁢Q⁢R⁢SPQRSPQRS is a rhombus. Since ∠⁢P⁢Q⁢Rnormal-∠PQR\angle PQR is a right angle, it follows that P⁢Q⁢R⁢SPQRSPQRS is a square.